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Suppose 0.458 g of potassium chloride is dissolved in 150. ml of a 77.0 mm aqueous solution of ammonium sulfate.rnrncalculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the potassium chloride is dissolved in it.rnrnround your answer to 3 significant digits.
Accepted Answer
Step 1: Convert the mass of potassium chloride to moles.
moles of KCl = 0.458 g KCl * (1 mol KCl / 74.55 g KCl) = 0.00615 mol KCl
Step 2: Calculate the total volume of the solution in liters.
volume of solution (in L) = 150 mL * (1 L / 1000 mL) = 0.150 L
Step 3: Calculate the moles of chloride anion.
moles of Cl- = 0.00615 mol KCl * (1 mol Cl- / 1 mol KCl) = 0.00615 mol Cl-
Step 4: Calculate the final molarity of chloride anion.
molarity of Cl- = moles of Cl- / total volume of solution
molarity of Cl- = 0.00615 mol / 0.150 L
molarity of Cl- = 0.0410 M
Therefore, the final molarity of chloride anion in the solution is 0.0410 M.
moles of KCl = 0.458 g KCl * (1 mol KCl / 74.55 g KCl) = 0.00615 mol KCl
Step 2: Calculate the total volume of the solution in liters.
volume of solution (in L) = 150 mL * (1 L / 1000 mL) = 0.150 L
Step 3: Calculate the moles of chloride anion.
moles of Cl- = 0.00615 mol KCl * (1 mol Cl- / 1 mol KCl) = 0.00615 mol Cl-
Step 4: Calculate the final molarity of chloride anion.
molarity of Cl- = moles of Cl- / total volume of solution
molarity of Cl- = 0.00615 mol / 0.150 L
molarity of Cl- = 0.0410 M
Therefore, the final molarity of chloride anion in the solution is 0.0410 M.
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